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प्रश्न
Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + … to n terms
उत्तर
Given series
⇒ (33 – 23) + (53 – 43) + (73 – 63) + ...
= (33 + 53 + 73 + …) – (23 + 43 + 63 + …)
⇒ [33 + 53 + 73 + … (2n + 1)3] – [23 + 43 + 63 + … (2n)3]
∴ Tn = (2n + 1)3 – (2n)3
= (2n + 1 – 2n) [2n + 1)2 + (2n + 1)(2n) + (2n)2] ....[∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= 1 · [4n2 + 1 + 4n + 4n2 + 2n + 4n2]
= 12n2 + 6n + 1
Sn = `sum "T"_n = 12 sum n^2 + 6 sum n + n`
= `12 * (n(n + 1)(2n + 1))/6 + (6n(n + 1))/2 + n`
= 2n(n + 1)(2n + 1) + 3n(n + 1) + n
= n[2(n + 1)(2n + 1) + 3(n + 1) + 1]
= n[2(2n2 + 3n + 1) + 3n + 3 + 1]
= n[4n2 + 6n + 2 + 3n + 4]
= n[4n2 + 9n + 6]
= 4n3 + 9n2 + 6n
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